Question

For $x>0,$ if $f\left(x\right)={\int }_1^x\frac{{\log }_{e}t}{(1+t)}dt$ then $f\left(e\right)+f\left(\frac{1}{e}\right)$ is equal to:

• A. $\frac{1}{2}$
• B. $-1$
• C. $1$
• D. $0$

Answer 1

The correct option is A $\frac{1}{2}$

$f\left(e\right)+f\left(\frac{1}{e}\right)={\int }_1^e\frac{\ln t}{1+t}dt+{\int }_1^{1/e}\frac{\ln t}{1+t}dt=l_1+l_2$
$I_2={\int }_1^{\frac{1}{e}}\frac{\ln t}{1+t}dt$
Put $t=\frac{1}{z}\Rightarrow dt=-\frac{dz}{z^2}$
$I_2={\int }_1^e-\frac{\ln z}{1+\frac{1}{z}}\times (-\frac{dz}{z^2})={\int }_1^e\frac{\ln z}{z(z+1)}dz={\int }_1^e\frac{\ln t}{t(t+1)}dt$

$\therefore f\left(e\right)+f\left(\frac{1}{e}\right)={\int }_1^e\frac{\ln t}{1+t}dt+{\int }_1^e\frac{\ln t}{t(t+1)}dt$
$={\int }_1^e\frac{\ln t}{t}dt$
let $\ln t=u\Rightarrow \frac{1}{t}dt=du$
$={\int }_0^{1}udu=\frac{u^2}{2}{|}_0^1=\frac{1}{2}$